Question

If one end of the diameter is $(1,1)$ and the other end lies on the line $x+y=3$ , then locus of centre of circle is

• A. $x+y=1$
• B. $2(x-y)=5$
• C. $2x+2y=5$
• D. None of these

Answer 1

Answer: (C)

Let the other end be $(t,3-t)$ So, the equation of the variable circle is
$(x-1)(x-t)+(y-1)(y-3+t)=0$
$\Rightarrow x^2+y^2-(1+t)x-(4-t)y+3=0$
$\therefore$ The centre $(\alpha ,\beta )$ is given by
$\alpha =\frac{1+t}{2},\beta =\frac{4-t}{2}$
$\Rightarrow 2\alpha +2\beta =5$
Hence, the locus is $2x+2y=5$